The 90% confidence, the mean farm size for this region is between 822.16 and 1259.99 acres.
1. Calculate the sample mean and standard deviation:
Sample mean (m) = sum of all farm sizes / number of farms = 52054 / 50 = 1041.08 acres
Sample standard deviation (s) can be calculated using a statistical calculator or software. The result is approximately 581.54 acres.
2. Determine the critical t-score for 90% confidence:
Degrees of freedom (df) = n - 1 = 50 - 1 = 49
Using a t-distribution table with 49 df and 90% confidence, the critical t-score is approximately 1.676.
3. Calculate the margin of error:
Margin of error (ME) = critical t-score * standard deviation / square root of sample size
ME = 1.676 * 581.54 / sqrt(50) ≈ 218.92 acres
4. Construct the confidence interval:
Lower limit = sample mean - margin of error = 1041.08 - 218.92 ≈ 822.16 acres
Upper limit = sample mean + margin of error = 1041.08 + 218.92 ≈ 1259.99 acres
Therefore, with 90% confidence, the mean farm size for this region is between 822.16 and 1259.99 acres.
Question
Assume that farm sizes in a particular region are normally distributed with a population standard deviation of 150 acres. A random sample of 50 farm sizes in this region is given below in acres. Estimate the mean farm size for this region with 90% confidence. Round your answers to two decimal places and use ascending order.
Size-202,1092,1321,319,331,938,73,200,200,229,160,486,370,256,351,770,170,288,177,165,67,194,353,296,293,2179,934,1490,107,79,1756,206,171,1312,188,438,474,131,56,204,1397,162,537,608,171,181,409,175,210,2612